The problem asks us to find the value of $a$ in the quadratic equation $2x2 - 3x + a = 0$, given that its roots are in the ratio $1 : 2$.
Using Vieta's Formulas
Let the roots of the quadratic equation $Ax2 + Bx + C = 0$ be $p$ and $q$. According to Vieta's formulas:
- Sum of roots: $p + q = -B/A$
- Product of roots: $p * q = C/A$
In our equation, $2x2 - 3x + a = 0$, we have $A = 2$, $B = -3$, and $C = a$.
Applying the Ratio of Roots
We are given that the roots are in the ratio $1 : 2$. Let the roots be $k$ and $2k$.
Sum of roots:
$k + 2k = -(-3)/2$
$3k = 3/2$
Solving for $k$:
$k = (3/2) / 3$
$k = 1/2$
Product of roots:
$k * 2k = a/2$
$2k2 = a/2$
Calculating the Value of 'a'
Now, substitute the value of $k = 1/2$ into the product of roots equation:
$2 * (1/2)2 = a/2$
$2 * (1/4) = a/2$
$1/2 = a/2$
Multiply both sides by 2:
$a = 1$
Therefore, the value of $a$ is 1.
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Important Questions from Quadratic Equation
If 2x 2+ 5x + 1 = 0, then one of the values of \(x - \frac{1}{{2x}}\) is:
If \(a-\frac{12}{a}=1\) , where a > 0, then the value of \(a^2+\frac{16}{a^2}\) is:
If x 2 – 3x + 1 = 0, then the value of \(\frac{(x^4+\frac{1}{x^2})}{(x^2+5x+1)}\) is:
If \(\sqrt{x}{}-{1\over\sqrt{x}}=\sqrt5\) , \(x \ne 0\) , then what is the value of \((x^4+{1\over{x^2}})\over(x^2+1) \) ?
If x 2 + \(\frac{1}{x^2}\) = 18, x > 0, then find the value of x 3 + \(\frac{1}{x^3}\) .
